Problem: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{4x^2 + 16x + 12}{4x^2 + 48x + 108}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {4(x^2 + 4x + 3)} {4(x^2 + 12x + 27)} $ $ k = \dfrac{4}{4} \cdot \dfrac{x^2 + 4x + 3}{x^2 + 12x + 27} $ Simplify: $ k = \dfrac{x^2 + 4x + 3}{x^2 + 12x + 27}$ Next factor the numerator and denominator. $ k = \dfrac{(x + 3)(x + 1)}{(x + 3)(x + 9)}$ Assuming $x \neq -3$ , we can cancel the $x + 3$ $ k = \dfrac{x + 1}{x + 9}$ Therefore: $ k = \dfrac{ x + 1 }{ x + 9 }$, $x \neq -3$